What are the solutions to the equation 2x^2+12x=4 ?

The solutions to the equation 2x^2+12x=4 are x=-3+sqrt(11),x=-3-sqrt(11)

Find the zeros of 2x^2+12x=4

The zeros of 2x^2+12x=4 are x=-3+sqrt(11),x=-3-sqrt(11)

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2 x^{2} + 12 x = 4

x = - 3 + 11, x = - 3 - 11

Decimal Notation

x = 0.31662 \dots, x = - 6.31662 \dots

Solution steps

2 x^{2} + 12 x = 4

Move

4

to the left side

2 x^{2} + 12 x - 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 2 ( - 4 )}{2 \cdot 2}

1 2^{2} - 4 \cdot 2 (- 4) = 411

x_{1, 2} = \frac{- 12 \pm 4 11}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 12 + 4 11}{2 \cdot 2}, x_{2} = \frac{- 12 - 4 11}{2 \cdot 2}

x = \frac{- 12 + 4 11}{2 \cdot 2} : - 3 + 11

x = \frac{- 12 - 4 11}{2 \cdot 2} : - 3 - 11

The solutions to the quadratic equation are:

x = - 3 + 11, x = - 3 - 11

6 x^{2} - 7 x - 3 = 0

4 (x + 1)^{2} = 9

4 x^{2} + 268 = - 72 x

a \times a + a = 240

x^{2} - 14 x - 20 = 0

What are the solutions to the equation 2x^2+12x=4 ?
The solutions to the equation 2x^2+12x=4 are x=-3+sqrt(11),x=-3-sqrt(11)
Find the zeros of 2x^2+12x=4
The zeros of 2x^2+12x=4 are x=-3+sqrt(11),x=-3-sqrt(11)